What is the derivative of g(x)= 1/(5x^2+4x) ?

The derivative of g(x)= 1/(5x^2+4x) is -(10x+4)/((5x^2+4x)^2)

What is the first derivative of g(x)= 1/(5x^2+4x) ?

The first derivative of g(x)= 1/(5x^2+4x) is -(10x+4)/((5x^2+4x)^2)

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derivative of

g (x) = \frac{1}{5 x ^{2} + 4 x}

- \frac{10 x + 4}{( 5 x ^{2} + 4 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{5 x ^{2} + 4 x})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((5 x^{2} + 4 x)^{- 1})

Apply the chain rule:

- \frac{1}{( 5 x ^{2} + 4 x ) ^{2}} \frac{d}{d x} (5 x^{2} + 4 x)

= - \frac{1}{( 5 x ^{2} + 4 x ) ^{2}} \frac{d}{d x} (5 x^{2} + 4 x)

\frac{d}{d x} (5 x^{2} + 4 x) = 10 x + 4

= - \frac{1}{( 5 x ^{2} + 4 x ) ^{2}} (10 x + 4)

Simplify

- \frac{1}{( 5 x ^{2} + 4 x ) ^{2}} (10 x + 4) : - \frac{10 x + 4}{( 5 x ^{2} + 4 x ) ^{2}}

= - \frac{10 x + 4}{( 5 x ^{2} + 4 x ) ^{2}}

\frac{ln ( x )}{3 x}, \frac{( ln ( x ) ) ^{2}}{3 x}

y^{^{'}} - \frac{3}{x} y = \frac{y ^{5}}{x ^{6}}

\int_{3}^{\infty} e^{- \frac{x}{2}} d x

\int \frac{1}{( e ^{x} + e ^{- x} ) ^{2}} d x

\frac{d}{d x} ((arctan (4 x))^{2})

What is the derivative of g(x)= 1/(5x^2+4x) ?
The derivative of g(x)= 1/(5x^2+4x) is -(10x+4)/((5x^2+4x)^2)
What is the first derivative of g(x)= 1/(5x^2+4x) ?
The first derivative of g(x)= 1/(5x^2+4x) is -(10x+4)/((5x^2+4x)^2)